Solving the indefinite integral $\int \frac{1}{x^2+x+1} dx$

Question

I want to solve the following indefinite integral: (1) $$\int \frac{1}{x^2+x+1} dx$$ completing the square: (2) $$=\int \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx$$ Substitution: (3) $$u=\frac{2x+1}{\sqrt 3}$$ brings: (4) $$dx=\frac{\sqrt 3}{2} du$$ (5) $$= \frac{2}{\sqrt 3}\int \frac{1}{u^2+1} du$$ Then getting standard integral (6) $$= \arctan(u)$$ and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the $\sqrt 3$ beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?

Answer 1

$$\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\left(\frac{2x+1}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2=\frac{3}{4}\left(\left(\frac{2x+1}{\sqrt3}\right)^2+1\right)=\frac{3}{4}(u^2+1).$$ Since $\frac{2x+1}{\sqrt3}=u$, we have $$\left(\frac{2x+1}{\sqrt3}\right)'dx=du$$ or $$\frac{2}{\sqrt3}dx=du$$ or $$dx=\frac{\sqrt3}{2}du.$$ Thus, $$\int\frac{1}{x^2+x+1}dx=\int\frac{1}{\frac{3}{4}(u^2+1)}\cdot\frac{\sqrt3}{2}du=\frac{2}{\sqrt3}\int\frac{1}{1+u^2}du.$$